[issue26283] zipfile can not handle the path build by os.path.join()

[Ezio Melotti]

Ezio Melotti added the comment:

I don't think this is a bug.  The ZIP format specification requires the use of 
forward slashes[0]:

   4.4.17 file name: (Variable)

   4.4.17.1 The name of the file, with optional relative path.
   The path stored MUST not contain a drive or
   device letter, or a leading slash.  All slashes
   MUST be forward slashes '/' as opposed to
   backwards slashes '\' for compatibility with Amiga
   and UNIX file systems etc.

os.path.join() will use different path separators depending on the system.  If 
you don't want to hardcode the slashes in a string literal, you can simply use 
'/'.join(...) instead of os.path.join().

[0]: https://pkware.cachefly.net/webdocs/casestudies/APPNOTE.TXT

--
nosy: +ezio.melotti
resolution:  -> not a bug
stage:  -> resolved
status: open -> closed

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[issue26283] zipfile can not handle the path build by os.path.join()

[Anish Shah]

Changes by Anish Shah :


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nosy:  -anish.shah

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[issue26283] zipfile can not handle the path build by os.path.join()

[SilentGhost]

Changes by SilentGhost :


--
nosy: +alanmcintyre, serhiy.storchaka, twouters
versions: +Python 3.6 -Python 3.4

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[issue26283] zipfile can not handle the path build by os.path.join()

[Anish Shah]

Changes by Anish Shah :


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nosy: +anish.shah

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[issue26283] zipfile can not handle the path build by os.path.join()

[張伯誠]

New submission from 張伯誠:

I think the built-in library zipfile.py does not handle correctly the path 
build by os.path.join().

For example, assuming we have a zipfile named Test.zip which contanins a member 
xml/hello.xml,
if you want to extract the member out, then you hava to use 'xml/hello.xml', if 
using os.path.join('xml','hello.xml'), you will get an error. 

Platform: Windows7, Python3.4

>>> import zipfile,os
>>> f=zipfile.ZipFile("Test.zip",'r')
>>> f.extract('xml/hello.xml','.') # OK.
>>> f.extract(os.path.join('xml','hello.xml'),'.') # does not work.

If we fixed the zipfile.py, inside the method getinfo(self,name) of class 
ZipFile:

before:
def getinfo(self, name):
"""Return the instance of ZipInfo given 'name'."""  
info = self.NameToInfo.get(name)

if info is None:
raise KeyError(
'There is no item named %r in the archive' % name)

return info

after:
def getinfo(self, name):
"""Return the instance of ZipInfo given 'name'."""
if os.sep=='\\' and os.sep in name:
name=name.replace('\\','/')

info = self.NameToInfo.get(name)
if info is None:
raise KeyError(
'There is no item named %r in the archive' % name)

return info

Then this line work!
>>>  f.extract(os.path.join('xml','hello.xml'),'.') # OK!

of course, this line also work:
>>> f.extract('xml/hello.xml','.') # also OK!


I think it is a bug. Why?
Let's we take a closer look at the method:

info = self.NameToInfo.get(name)

The keys of NameToInfo are always the format "xxx/yyy" according to the 
document of class ZipInfo:

def __init__(self, filename="NoName", date_time=(1980,1,1,0,0,0)):
 
# This is used to ensure paths in generated ZIP files always use
# forward slashes as the directory separator, as required by the
# ZIP format specification.

if os.sep != "/" and os.sep in filename:

filename = filename.replace(os.sep, "/")

Hence the method getinfo(self,name) of class ZipFile always get KeyError for 
the path build os.path.join('xxx','yyy')


Thnaks for reading!
Bocheng.

--
components: Library (Lib)
messages: 259524
nosy: 張伯誠
priority: normal
severity: normal
status: open
title: zipfile can not handle the path build by os.path.join()
type: behavior
versions: Python 3.4

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